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4x^2+x=1
We move all terms to the left:
4x^2+x-(1)=0
a = 4; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·4·(-1)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*4}=\frac{-1-\sqrt{17}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*4}=\frac{-1+\sqrt{17}}{8} $
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